Problem: The lifespans of zebras in a particular zoo are normally distributed. The average zebra lives $20.5$ years; the standard deviation is $3.9$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a zebra living less than $32.2$ years.
$20.5$ $16.6$ $24.4$ $12.7$ $28.3$ $8.8$ $32.2$ $99.7\%$ $0.15\%$ $0.15\%$ We know the lifespans are normally distributed with an average lifespan of $20.5$ years. We know the standard deviation is $3.9$ years, so one standard deviation below the mean is $16.6$ years and one standard deviation above the mean is $24.4$ years. Two standard deviations below the mean is $12.7$ years and two standard deviations above the mean is $28.3$ years. Three standard deviations below the mean is $8.8$ years and three standard deviations above the mean is $32.2$ years. We are interested in the probability of a zebra living less than $32.2$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the zebras will have lifespans within 3 standard deviations of the average lifespan. The remaining $0.3\%$ of the zebras will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({0.15\%})$ will live less than $8.8$ years and the other half $({0.15\%})$ will live longer than $32.2$ years. The probability of a particular zebra living less than $32.2$ years is ${99.7\%} + {0.15\%}$, or $99.85\%$.